Class 8 Maths Notes for Algebraic Expressions And Identities

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CLASS 8
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ALGEBRAIC EXPRESSION AND IDENTITIES
Important Concepts/ Result
1. Terms are formed by the product of variables and constants, e.g. –3xy, 2xyz, 5x, etc.
2. Terms are added to form expressions, e.g. –2xy + 5x.
3. Expressions that contain exactly one, two and three terms are called monomials, binomials and
trinomials, respectively.
4. In general, any expression containing one or more terms with non– zero coefficients (and with
variables having non–negative exponents) is called a polynomial.
5. Like terms are formed from the same variables and the powers of these variables are also the same.
But coefficients of like terms need not be the same.
6. There are number of situations like finding the area of rectangle triangle, etc. in which we need to
multiply algebraic expressions.
7. Multiplication of two algebraic expressions is again an algebraic expression.
8. A monomial multiplied by a monomial always gives a monomial.
9. While multiplying a polynomial by a monomial, we multiply every term in the polynomial by the
monomial using the distributive law a (b+c) = ab + ac.
10. In the multiplication of a polynomial by a binomial (or trinomial), we multiply term by term, i.e.
every term of the polynomial is multiplied by every term in the binomial (or trinomial) using the
distributive property.
MCQs (illustrations/Examples)
1. Subtract the algebraic expressions: 6a – 3b and 4a + 2b.
A. 2𝑎 – 5𝑏
B. 2𝑎 + 5𝑏
C. 10𝑎 – 5𝑏
D. 10𝑎 + 5𝑏
Answer: 𝐵. 2𝑎 + 5𝑏
2. Evaluate the product using the distributive property: (5𝑥 + 3𝑦)(4𝑥 – 2𝑦).
A. 20x2 – 6xy
B. 20x2 + 18xy
C. 20x2 – 10xy + 6y2
D. 20x2 + 22xy – 6y2
Answer: C. 20x2 – 10xy + 6y2
3. Multiply the trinomials: (𝑥 + 2𝑦 + 3𝑧)(𝑥 – 2𝑦 – 3𝑧).
A. x2 – 4y2 – 9z2
B. x2 + 4y2 + 9z2
C. x2 – 4y2 + 9z2
D. x2 + 4y2 – 9z2
Answer: A. x2 – 4y2 – 9z2
4. Evaluate the product: (2x + y)(x2 – xy + y2).
A. 2x3 + 3x2y – 2xy2 + y3
B. 2x3 – x2y + xy2 + y3
C. 2x3 + x2y – 3xy2 + y3
D. 2x3 – x2y + 2xy2 + y3
Answer: B. 2x3 – x2y + xy2 + y3

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Case Study

5 In a small town, there is a group of students from 'Creative Minds' school. The Mathematics teacher,
Mr. Sharma, has given students a challenge to find the total area of three unique gardens A, B, and C, and to
compare these three gardens to determine which one has the largest area. To complete this challenge, the
students need to apply their knowledge of algebraic expressions and their skills of addition, subtraction,
multiplication, and simplification.
The area of each garden is represented as follows:
Garden A: (x2 + 3x) sq. meters
Garden B: (2x2 – 7x) sq. meters
Garden C: (1/2 x2 + 5x + 8) sq. Meters

Question 1: Find the sum of algebraic expressions representing the area of Garden A and Garden B.
Solution 1:
Sum of the area of Garden A and Garden B
= (x2 + 3x) + (2x2 – 7x)
= x2 + 3x + 2x2 – 7x
= (x2 + 2x2) + (3x – 7x)
= 3x2 – 4x

Question 2: Simplify the algebraic expression representing the area of Garden C.
Solution 2:
Area of Garden C = 1/2 x2 + 5x + 8
Since there are no like terms present, this expression is already simplified.

Question 3: Find the product of the area of Garden A and Garden B.
Solution 3:
Product of area of Garden A and Garden B = x2 + 3x 2x2 – 7x
Now we need to multiply each term in the first expression by each term in the second expression:
a. Monomial multiplied by a monomial = x2 X 2x2 = 2x4
b. Monomial multiplied by a binomial or trinomial (expanding brackets):
= x2 X (–7x) + 3x X (2x2) + 3x X (–7x)
= –7x3 + 6x3 – 21x2
c. Combine the results from steps a and b and simplify the expression:
= 2x4 + (–7x3 + 6x3) – 21x2
= 2x4 – x3 – 21x2
The product of the area of Garden A and Garden B is 2x4 – x3 – 21x2.

Short answer type question

Q1. If A = 5x² – 3x + 2 and B = 2x² + 4x – 5, find the sum and the difference of A and B.
Solution:
𝑆𝑢𝑚: 𝐴 + 𝐵 = (5𝑥² – 3𝑥 + 2) + (2𝑥² + 4𝑥 – 5)
= (5𝑥² + 2𝑥²) + (– 3𝑥 + 4𝑥) + (2 – 5) = 7𝑥² + 𝑥 – 3

𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒: 𝐴 – 𝐵 = (5𝑥² – 3𝑥 + 2) – (2𝑥² + 4𝑥 – 5)
= (5𝑥² – 2𝑥²) + (– 3𝑥 – 4𝑥) + (2 + 5) = 3𝑥² – 7𝑥 + 7

2. Simplify the following expression using the distributive property: 2(x + 3) + 5(2x – 1)
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Solution:
Using the distributive property, we get:
= 2𝑥 + 6 + 10𝑥 – 5
= 12𝑥 + 1
Therefore, the simplified expression is 12x + 1.

3 Find the product of two trinomials – (2x + y – 3z) and (x + 2y + z) using the distributive property.
Solution:
=(2x + y – 3z)(x + 2y + z)
= (2x)(x) + (2x)(2y) + (2x)(z) + (y)(x) + (y)(2y) + (y)(z) – (3z)(x) – (3z)(2y) – (3z)(z)
= 2x2 + 4xy + 2xz + xy + 2y2 + yz – 3zx – 6zy – 3z2
= 2x2 + 5xy + 2y2 – zx – 6zy +2xz + yz – 3z2

Long answer type questions
1 In the given expressions, (A) 2𝑥 2 – 5𝑥 + 1 and (B) 7x2 – 11x + 2, add and subtract the expressions,
and simplify the resulting expressions. Furthermore, multiply expression (A) with expression (B) using the
distributive property and simplify.
Solution:
Step 1: Addition of algebraic expressions.
(2x2 – 5x + 1) + (7x2 – 11x + 2)
= 2x2 – 5x + 1 + 7x2 – 11x + 2 (Combine like terms.)
= (2x2 + 7x2) + (–5x – 11x) + (1 + 2)
= 9x2 – 16x + 3

Step 2: Subtraction of algebraic expressions.
(2x2 – 5x + 1) – (7x2 – 11x + 2)
= 2x2 – 5x + 1 – 7x2 + 11x – 2 (Combine like terms.)
= (2x2 – 7x2) + (–5x + 11x) + (1 – 2)
= –5x2 + 6x – 1

Step 3: Multiplication of algebraic expressions using the distributive property.
(2x2 – 5x + 1)(7x2 – 11x + 2)
= 2x2(7x2) + 2x2(–11x) + 2x2(2) – 5x(7x2) – 5x(–11x) – 5x(2) + 1(7x2) – 1(11x) + 1(2)
= 14x4 – 22x3 + 4x2 – 35x3 + 55x2 – 10x + 7x2 – 11x + 2
= 14x4 + (–22x3 – 35x3) + (4x2 + 55x2 + 7x2) + (–10x –11x)
= 14x4 – 57x3 + 66x2 – 21x + 2

The added expression is 9x2 – 16x + 3. The subtracted expression is –5x2 + 6x –1. The product of the
expressions using the distributive property is 14x4 – 57x3 + 66x2 – 21x + 2.

2 Consider the following algebraic expressions: expression (A) 3x3 – 5x2 + 4x – 6 and expression (B)
2
2x – 3x + 1. Find the product of expression (A) by the monomial 2y, and expression (B) by the binomial (x
– 4y). Use the distributive property to expand the product.
Solution:
Step 1: Product of expression (A) by the monomial 2y.
(3x3 – 5x2 + 4x – 6)(2y)
= 3x3(2y) – 5x2(2y) + 4x(2y) – 6(2y)
= 6x3y – 10x2y + 8xy – 12y
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Step 2: Product of expression (B) by the binomial (x – 4y).
(2x2 – 3x + 1)(x – 4y)
= 2x2 x + 2x2 –4y – 3x x – 3x –4y + 1 x + 1 –4y
= 2x3 – 8x2y – 3x2 + 12xy + x – 4y

The product of expression (A) and the monomial 2y is 6x3y – 10x2y + 8xy – 12y. The product of
expression (B) and the binomial (x – 4y) is 2x3 – 8x2y – 3x2 + 12xy + x – 4y.

MCQs (Questions for Practice)
1. What is the sum of the algebraic expressions: 3x + 5y and 2x – 4y?
𝐴. 1𝑥 + 1𝑦 𝐶. 5𝑥 – 1𝑦
𝐵. 5𝑥 + 1𝑦 𝐷. 5𝑥 + 9𝑦
2. Simplify the algebraic expression: 4(3m –2n) – 2(4m +5n).
A. –2m – 16n C. 2m + 16n
B. –2m + 16n D. 2m – 16n
3. Multiply the monomials: 4x2y X 3xy3
A. 12x2y4 C. 12x3y4
B. 7x3y3 D. 12x3y3
4. Multiply the monomial by the binomial: 3x(2x – 5).
A. 6x2 – 15x C. 6x2 + 5x
B. 5x2 – 15x D. 5x2 + 15x

5. Multiply the binomials using the distributive property: (a + 5)(2a – 3).
A. 2a2 – 3a + 10 C. 2a2 + 10a – 15
B. 2a2 + 7a + 15 D. 2a2 – 3a +10a – 15
6. Simplify the expression: (x2 + 3x – 2)(x – 1).
A. x3 + 2x2 – 5x + 2 C. x2 + 2x2 – 5x – 2
B. x3 + 2x2 + 5x – 2 D. x2 + 2x2 + 5x + 2
7. Expand the binomial expression: (2x + 3)(3x – 4).
A. 6x2 – 5x – 12 C. 6x2 + x – 12
2
B. 6x – 8x + 9x – 12 D. 6x2 – 3x + 12
8. Expand and simplify the expression: (2m – n)(2m + n).
A. 4m2 – n2 C. 4m2 – 2mn + n2
B. 4m2 + n2 D. 4m2 + 2mn – n2
2 2
9. Evaluate the expression: (4x – 3xy + 2y )(2x + y).
A. 8x3 – x2y + 2xy2 C. 8x3 – x2y – xy2
B. 8x3 + 5x2y – xy2 D. 8x3 + 6x2y + xy2
10. Simplify the expression: (3a – 2b)(a2 + 2ab + b2).
A. 3a3 – 6a2b + 3ab2 +2ab3
B. 3a3 – 6a2b + 4b3
C. 3a3 – 4b3
D. 3a3 – 2a2b + 2b3

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Short answer type question
1. Add the following algebraic expressions: 2𝑥 + 5𝑦, 3𝑥 – 4𝑦, – 𝑥 + 6𝑦

2. Subtract the following algebraic expressions: (3𝑎 – 4𝑏) – (2𝑎 + 5𝑏)

3. Multiply the following monomials: 4x and 5x

4. Multiply the following monomials: –3a²b and 4ab²

5. Multiply the following algebraic expressions: 2𝑥(3𝑥 – 4𝑦)

6. Multiply the following algebraic expressions: 5a(2a2 – 3ab + b2)

7. Multiply the following binomials: (𝑥 + 3)(𝑥 – 2)

8. Multiply the following binomials: (2x – 5)(x + 4)

9. Multiply the following trinomials: (x2 + 2x + 1)(x + 1)
10. Simplify (a + b) (2a – 3b + c) – (2a – 3b) c.

Long answer type questions
1 Consider two algebraic expressions: A = 4𝑥 2 – 5𝑥𝑦 + 7𝑦 2 and 𝐵 = – 3𝑥 2 + 2𝑥𝑦 – 9𝑦 2 . Find the
sum and difference of these two algebraic expressions.
2 Given two monomials: M1 = 3𝑥 2 𝑦 and M2 = – 5𝑥𝑦 2 . Calculate the product of M1 and M2.
3 Compute the product of the monomial 2𝑥 2 with the binomial (3𝑥 – 4𝑦).
4 Calculate the product of the monomial (–x) with the trinomial 6𝑥 2 – 8𝑥𝑦 + 9𝑦 2 .
5 Multiply the following pair of binomials: 4𝑥 – 3𝑦 and 2𝑥 + 5𝑦 .
6 Find the product of two trinomials: 𝑥 2 – 2𝑥 + 1 and 𝑥 2 – 4𝑥 – 3 .
Answers: MCQs (Questions for Practice)

1. 5x + 1y
2. –2m – 16n
3. 12x3y4
4. 6x2 – 15x
5. 2a2 – 3a +10a – 15
6. x3 + 2x2 – 5x + 2
7. 6x2 – 8x + 9x – 12
8. 4m2 – n2
9. 8x3 + 5x2y – xy2
10. 3a3 – 6a2b + 4b3

Answer Key: Short answer type questions
1. 4x + 7y
2. a – 9b
3. 20x2
39

4. –12a³b³
5. 6x2 – 8xy
6. 10a3 – 15a2b + 5ab2
7. x2 + x – 6
8. 2x2 + 3x – 20
9. x3 + 3x2 + 3x + 1
10. 2a2–3b2–ab+4bc–ac

Answer Key: Long answer type questions
1. 𝑆𝑢𝑚 = 1𝑥 2 – 3𝑥𝑦 – 2𝑦 2
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 7𝑥 2 – 7𝑥𝑦 + 16𝑦 2
2. – 15𝑥 3 𝑦 3
3. 6𝑥 3 – 8𝑥 2 𝑦
4. – 6𝑥 3 + 8𝑥 2 𝑦– 9𝑥𝑦 2
5. 8𝑥 2 + 14𝑥𝑦 – 15𝑦 2
6. 𝑥 4 – 6𝑥 3 + 6𝑥 2 + 6𝑥 – 3
Test 1 (Total Marks: 20)
Q1. List the coefficients and variables/variables in the following expression: (2marks)
a) -4x³ + 2x²y - 5xy² + 3y³
b) 6z² - 8yz + 5xz³ - 10y³z²
Q2. Find the product of the following expressions: (2 marks)
a) (x - 3y)(x + 2y)
b) (3a + 5)(3a - 5)
Q3. Identify the coefficients of the terms in the following algebraic expressions. (2marks)
a) 3x - 4y + 5z
b) 5x²y - 7xy⁴ + 11y³
Q4. Evaluate the expressions below when a = 3, b = -2, and c = 4. (4marks)
a) 3a² + 5c - b
b) a² - b³ + 3c²
c) (2a + c)(a - 2b) - b²
d) 2(3a - 5b) + ½(4c - 10a)
Q5. Simplify the following expressions: (3)
a) 3x - 7y + 2x + y
b) (2a - 3b) + (-5a + 4b)
c) ½ (8m - 12n) - ⅓ (9m + 18n)
Q6. Determine whether the following pairs are like terms or unlike terms: (3 marks)
a) 5xy, -2yx
b) 6xyz, 6x²y
c) 3ab², 3b²a
Q7 Write the algebraic expressions for the following statements. (4 marks)
a) x is multiplied by 2 and then added to 5
b) The difference between twice y and 3
c) x is multiplied by the sum of y and 7
d) z is divided by 3 and subtracted from the product of 5 and x

Test 2 (30 marks)
40

Objective Type Questions (1 mark each)
1. Simplify the expression: 3x + 5 - 2x
a) x + 5 c) 5x
b) x - 2 d) 5x + 5

2. The coefficient of x2 in the term 4x3y5 – 5x2y2 is:
a) -5 c) 2
b) 4 d) 5

3. Which of the following is a like term for 5x2y?
a) 7x2y c) 5x3y
b) 5x2y2 d) 7xy

Multiple Choice Questions (2 marks each)
4. Express -5x^3 – 10x^2 + 15 as a product of its factors.
a) −5𝑥 𝑥 3 + 2𝑥 2 – 3 c) −5𝑥 𝑥 3 – 2𝑥 2 + 5
b) −5𝑥 𝑥 2 + 2𝑥 – 5 d) 5𝑥 𝑥 3 – 2𝑥 2 – 3

5. Find the product of (2𝑥 − 3𝑦)(𝑥 + 𝑦)
a) 2𝑥 2 – 𝑥𝑦 – 3𝑦 2
b) 2𝑥 2 – 3𝑥𝑦 + 3𝑦 2
c) 2𝑥 2 + 𝑥𝑦 – 3𝑦 2
d) 2𝑥 2 – 𝑥𝑦 + 3𝑦 2

Short Answer Type Questions (3 marks each)
6. If A = 2𝑥 𝑦 – 5𝑥 𝑦 + 𝑥 𝑦 and B = 𝑥 2 𝑦 – 3𝑥 3 𝑦 2 + 4𝑥 𝑦 2 , find A + B.
3 2 2 2

7. Solve for x: 4𝑥 – 3 2𝑥 – 1 = 5𝑥 + 6

Long Answer Type Questions (4 marks each)
8. Simplify the expression: 2(x - y)2 + 3(x2 - y2) - (x2 - 2xy + y2) and find its value when x = -1 and y
= 2.

9. Karan received 200 coins consisting of x nickels and (200 - x) dimes. If the total amount of money
he received was Rs 18, write an algebraic expression to represent this situation and find the number
of nickels.

Case Study Based Question (4 marks)
11. A factory is packaging instant noodles in two types of boxes. Box A contains 3 packages of chicken-
flavored noodles, 2 packages of beef-flavored noodles, and 1 package of vegetable-flavored noodles.
Box B contains 5 packages of chicken-flavored noodles and 1 package of beef-flavored noodles. If
the total number of packages of chicken-flavored noodles is C, packages of beef-flavored noodles is
B, and packages of vegetable-flavored noodles is V, write algebraic expressions for the total number
of packages in box A and box B. Then, find the total number of packages for each flavor if the
factory produced 100 box A's and 150 box B's.

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Answer key (TEST 1)
Q1.
a) Coefficients: -4, 2, -5, 3; Variables: x, y
b) Coefficients: 6, -8, 5, -10; Variables: z, y, x

Q2.
a) x² - 3yx + 2yx - 6y² = x² - y²
b) (9a² - 15a + 15a - 25) = 9a² - 25

Q3.
a) Coefficients: 3, -4, 5
b) Coefficients: 5, -7, 11
Q4.
𝑎) 3(3)² + 5(4) − (−2) = 3(9) + 20 + 2 = 27 + 22 = 49
𝑏) (3)² − (−2)³ + 3(4)² = 9 − (−8) + 3(16) = 9 + 8 + 48 = 65
𝑐) (2(3) + 4)(3 − 2(−2)) − (−2)² = (6 + 4)(3 + 4) − 4 = 10(7) − 4 = 70 − 4
= 66
𝑑) 2(3(3) − 5(−2)) + ½(4(4) − 10(3)) = 2(9 + 10) + ½(16 − 30) = 2(19) − 7
= 38 − 7 = 31
Q5.
𝑎) 3𝑥 − 7𝑦 + 2𝑥 + 𝑦 = (3𝑥 + 2𝑥) + (−7𝑦 + 𝑦) = 5𝑥 − 6𝑦
𝑏) (2𝑎 − 3𝑏) + (−5𝑎 + 4𝑏) = (2𝑎 − 5𝑎) + (−3𝑏 + 4𝑏) = −3𝑎 + 𝑏
𝑐) ½(8𝑚 − 12𝑛) − ⅓(9𝑚 + 18𝑛) = 4𝑚 − 6𝑛 − 3𝑚 − 6𝑛
= (4𝑚 − 3𝑚) + (−6𝑛 − 6𝑛) = 𝑚 − 12𝑛
Q6.
a) Like terms
b) Unlike terms
c) Like terms

Q7.
a) 2x + 5
b) 2y - 3
c) x(y + 7)
7
d) 5x – 3

Answer Key: (TEST 2)
1. 𝑎) 𝑥 + 5
2. 𝑎) − 5
3. 𝑎) 7𝑥 2 𝑦
4. b) −5𝑥 𝑥 2 + 2𝑥 – 5
5. a) 2𝑥 2 – 𝑥𝑦 – 3𝑦 2
6. −𝑥 3 𝑦 2 – 𝑥𝑦 2 + 𝑥 2 𝑦
7. x = 1/5
8. 5𝑥 2 – 4𝑥𝑦 + 5𝑦 2 , when x = -1 & y = 2; the value is 57
9. x = 160 (number of nickels)
10.Box A: 3C + 2B + V
Box B: 5C + B
- C = 3(100) + 5(150) = 900
- B = 2(100) + 1(150) = 350
- V = 1(100) = 100

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